Solution: Since $\angle DBA$ = $\angle FBC$ and $\angle BDA = \angle BFC = 90º$. So $\triangle BDA$ $\sim$ $\triangle BFC$. Since $T$ is the mid point of $DA$ and $S$ is the mid point of $FC$ so $\triangle DBT$ $\sim$ $\triangle FBS$ so $\angle DTB = \angle BSF$. Since $\angle BDA =\angle BDT = 90$ and $D$ is the mid point of $TN$ so $\angle BSF = \angle BTD =\angle BND$ so $BNHS$ is cyclic $(1)$. By symmetric so $BT = BN = BM$ so $B$ is the circumcenter of $\triangle NMT$ so $\angle NBM = 2\angle NTM =\angle HTM + \angle HMT = 180º - \angle THM=180º-\angle NHM$ so $NHMB$ cyclic $(2)$.From $(1),(2)$ so $BNHSM$ cyclic so $BMNS$ cyclic so $S$ lies on the circumcircle of triangle $BMN$.

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